设数列{an}的前n项和sn,已知a1=2,sn=an+2^n-2,设bn=log2an,求数列{bn}的前n项和Tn

Question

设数列{an}的前n项和sn,已知a1=2,sn=an+2^n-2,设bn=log2an,求数列{bn}的前n项和Tn

Answer 1

sn=an+2^n-2,
sn-1=an-1+2^(n-1)-2,
an=sn-sn-1
=an+2^n-2-[an-1+2^(n-1)-2]
=an+2^n-an-1-2^(n-1)
an-1=2^n-2^(n-1)
=2*2^(n-1)-2^(n-1)
=2^(n-1)
即an=2^n
bn=log2an
b1=log2(2^1)=1
b2=log2(2^2)=2
b3=log2(2^3)=3
.
bn=log2(2^n)=n
Tn=1+2+3+..+n=n(n+1)/2